Cuando se trata Ca3P2 con agua, los productos son Ca(OH)2 y PH3. Calcular el peso máximo obtenido al reaccionar 2 g de Ca3P2 con 1 g de H2O.

Answers

Answer 1

Answer:

0.629 gramos de PH3 es la máxima de cantidad que puede ser producida.

Explanation:

¡Hola!

En este caso, dado que tenemos la siguiente reacción química, la cual se puede balancear directamente:

[tex]Ca_3P_2+6H_2O\rightarrow 3Ca(OH)_2+2PH_3[/tex]

Podemos calcular la masa máxima de cualquier producto, digamos PH3, al comparar la masa de este, que 2 g the Ca3P2 y 1 g de H2O producen por separado y de acuerdo a la estequiometría:

[tex]2gCa_3P_2 *\frac{1molCa_3P_2 }{182.18gCa_3P_2 }*\frac{2molPH_3}{1molCa_3P_2 } *\frac{34gPH_3}{1molPH_3 } =0.747gPH_3\\\\1gH_2O *\frac{1molH_2O }{18.02gH_2O}*\frac{2molPH_3}{6molH_2O} *\frac{34gPH_3}{1molPH_3 } =0.629gPH_3[/tex]

De este modo, infermos que solamente 0.629 gramos the PH3 pueden ser obtenidos al ser el agua el reactivo límite.

¡Saludos!


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Answers

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